1. ## "problem of inequalities"

"problem of inequalities"

2. Please post only one question at a time.

$\sqrt{3x+1}

Square both sides:

$3x+1<(x+1)^2 =\Rightarrow 3x+1

$0

$0

$0>x, 1

Only one solution will work in this case

$1 < x \Rightarrow x > 1$

EDIT: DivideBy0 pointed out a spot that wouldn't work, which I suspected... grr

$3x+1 \ge 0$

$x \ge \frac{-1}{3}$

3. I think you made a little typo there which led to a mistake.

$x<0$ or $x > 1$

But, to be defined,

$3x+1\geq 0$

$3x\geq -1$

$x \geq -\frac{1}{3}$

So the solution set is $\{x:-\frac{1}{3}\leq x < 0 \cup x>1\}$

4. Originally Posted by afeasfaerw23231233
"problem of inequalities"
For the second question, look at the graph of that function.

Can you see it intercepts the axis only once?

So $x \geq 3$ satisfies the equation

5. $\frac{x-2}{x+2} > \frac{x^2-4}{x^2+4}$

$\left(\frac{x-2}{x+2}\right)\left(\frac{x-2}{x-2}\right)>\frac{x^2-4}{x^2+4}$

$\frac{(x-2)^2}{x^2-4}>\frac{x^2-4}{x^2+4}$

$\left(\frac{(x-2)^2}{x^2-4}\right)\left(\frac{x^2+4}{x^2+4}\right)>\frac{x^ 2-4}{x^2+4}\left(\frac{x^2-4}{x^2-4}\right)$

$\frac{(x^2+4)(x-2)^2}{x^4-16}>\frac{(x^2-4)^2}{x^4-16}$

Case 1

If $x^4-16>0\Rightarrow x > 2$ or $x<-2$,

$(x^2+4)(x-2)^2>(x^2-4)^2$

$x^4-4x^3+8x^2-16x+16>x^4-8x^2+16$

$4x^3-16x^2+16x<0$

$4x(x^2-4x+4)<0$

$4x(x-2)^2<0$

Which is true so long as $x<0$.

Case 2

If $x^4-16<0\Rightarrow -2 < x< 2$, after some simplification we arrive at

$4x(x-2)^2>0$

Which is true so long as $x>0$.

So the two regions where the inequality is true is when

$-2 AND $x>0$ $\Rightarrow 0.

Or

$x<-2 \cap x>2$ AND $x<0$ $\Rightarrow x<-2$.

So the possible values of x are $\{x:0